Practice Problems: Integation By Parts

Integration by parts is an integration technique that involves separating the integrand into two parts. This technique is useful whenever a product appears in the integrand and the implement u-substitution is not feasible.

\(\displaystyle\int u\,dv = uv\,-\!\int v\,du\)

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Practice Problems

Answer the following questions by using the integration technique known as integration by parts (IBP).

\(\displaystyle \int x\texttt{sin}(2x) dx \)
\(\displaystyle \int x^2\texttt{sin}(2x) dx \)
\(\displaystyle \int \texttt{arctan}(2x+1) dx \)
\(\displaystyle \int \texttt{cos}(\sqrt{x}) dx \)

\(\displaystyle \int \texttt{e}^{\sqrt{x}} dx \)
\(\displaystyle \int \texttt{sin}(\texttt{ln}(x)) dx \)
\(\displaystyle \int \texttt{e}^{2x}\texttt{cos}(5x) dx \)
\(\displaystyle \int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx \)

\(\displaystyle \int \texttt{e}^{\alpha x}\texttt{cos}(\beta x) dx \)

Suggestive Solution Guide

Problem 1: Solution

Let \(u=x\), so \(du=dx\); and \(dv=\texttt{sin}(2x)dx\), so \(v=\int\texttt{sin}(2x)dx=-\frac{1}{2}\texttt{cos}(2x)\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int x\texttt{sin}(2x) dx
&= \color{red}{\underbrace{\color{black}{x}}_{\large u}} \color{red}{\underbrace{ \color{black}{\left( \frac{-1}{2}\texttt{cos}(2x) \right)}}_{\large v}}\:- \int\ \color{red}{\underbrace{\color{black}{\left(\frac{-1}{2}\texttt{cos}(2x)\right)}}_{\large v}} \color{red}{\underbrace{\color{black}{dx}}_{\large du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{-x}{2}\texttt{cos}(2x)\:+ \frac{1}{2}\int \texttt{cos}(2x)dx
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= \color{green}{\frac{-x}{2}\texttt{cos}(2x)\:+ \frac{1}{4}\texttt{sin}(2x)\:+ C}
& \color{gray}{\text{[Integrate using a quick u-sub]}} \\[5pt]
\end{align*}
\)

Problem 2: Solution

Let \(u=x^2\), so \(du=2xdx\); and \(dv=\texttt{sin}(2x)dx\), so \(v = \int \texttt{sin}(2x)dx = -\frac{1}{2}\texttt{cos}(2x)\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int x^2\texttt{sin}(2x)dx
&= \color{red}{\underbrace{\color{black}{x^2}}_{\large u}} \color{red}{\underbrace{\color{black}{\left( \frac{-1}{2}\texttt{cos}(2x) \right)}}_{\large v}}\:- \int \color{red}{\underbrace{\color{black}{\left(\frac{-1}{2}\texttt{cos}(2x)\right)}}_{\large v}} \color{red}{\underbrace{\color{black}{(2xdx)}}_{\large du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= -\frac{x^2}{2}cos(2x)\:+ \int x\texttt{cos}(2x)dx
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
\end{align*}
\)

Apply IBP once again. Let \(u=x\), so \(du=dx\); and \(dv=\texttt{cos}(2x)dx\), so \(v = \int \texttt{cos}(2x)dx = \frac{1}{2}\texttt{sin}(2x)\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
-\frac{x^2}{2}\texttt{cos}(2x)\:+ \int x\texttt{cos}(2x)dx
&= -\frac{x^2}{2}\texttt{cos}(2x)\:+ \left(\color{red}{\underbrace{\color{black}{x}}_{\large u}} \color{red}{\underbrace{\color{black}{\left( \frac{1}{2}\texttt{sin}(2x) \right)}}_{\large v}}\:- \int \color{red}{\underbrace{\color{black}{\left(\frac{1}{2}\texttt{sin}(2x)\right)}}_{\large v}} \color{red}{\underbrace{\color{black}{(dx)}}_{\large du}}\right)
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= -\frac{x^2}{2}\texttt{cos}(2x)\:+ \frac{x}{2}\texttt{sin}(2x)\:- \frac{1}{2}\int \texttt{sin}(2x)dx
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= -\frac{x^2}{2}\texttt{cos}(2x)\:+ \frac{x}{2}\texttt{sin}(2x)\:- \left(\frac{1}{2}\right) \left(\frac{-1}{2}\texttt{cos}(2x)\right)\:+ C
& \color{gray}{\text{[Integrate using a quick u-sub]}} \\[5pt]
&= \color{green}{-\frac{x^2}{2}\texttt{cos}(2x)\:+ \frac{x}{2}\texttt{sin}(2x)\:+ \frac{1}{4}\texttt{cos}(2x)\:+ C}
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
\end{align*}
\)

Problem 3: Solution

Let \(u=\texttt{arctan}(2x+1)\), so \(du=\frac{2}{(2x+1)^2+1}dx\); and \(dv=dx\), so \(v=x\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{arctan}(2x+1)dx
&= \color{red}{\underbrace{\color{black}{\texttt{arctan}(2x+1)}}_{\large u}} \color{red}{\underbrace{\color{black}{x}}_{\large v}}\:- \int \color{red}{\underbrace{\color{black}{x}}_{\large v}} \color{red}{\underbrace{\color{black}{\left(\frac{2}{(2x+1)^2+1}\right)}}_{\large du}}dx
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{arctan}(2x+1)\:- 2\int \frac{x}{(2x+1)^2+1}dx
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
\end{align*}
\)

Use U-Sub with the given integrand. Let \(u=2x+1\), so \(du=2dx\); and \(x=\frac{u-1}{2}\). While working the only integral left, it’s fine to have \(u\) and \(x\) at the same time given that the integrand is expressed in regards of just one variable. Substituting accordingly yields:

\(\normalsize
\begin{align*}\displaystyle
x\texttt{arctan}(2x+1)\:- 2\int \frac{x}{(2x+1)^2+1}dx
&= x\texttt{arctan}(2x+1)\:- 2\int \left( \frac{\color{red}{\left( \frac{u-1}{2} \right)}}{\color{red}{u^2}+1} \right) \color{red}{\left( \frac{du}{2} \right)}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{arctan}(2x+1)\:- \frac{1}{2}\int\frac{u-1}{u^2+1}du
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= x\texttt{arctan}(2x+1)\:- \frac{1}{2}\left( \int\frac{u}{u^2+1}du\:-\int\frac{1}{u^2+1}du \right)
& \color{gray}{\text{[Split the integrals strategically]}} \\[5pt]
&= x\texttt{arctan}(2x+1)\:- \frac{1}{2}\int\frac{u}{u^2+1}du\:+\frac{1}{2}\int\frac{1}{u^2+1}du
& \color{gray}{\text{[Distribute the fraction]}} \\[5pt]
&= x\texttt{arctan}(2x+1)\:+ \frac{1}{2}\texttt{arctan}(u)\:- \frac{1}{2}\int\frac{u}{u^2+1}du
& \color{gray}{\text{[Integrate the second integral]}} \\[5pt]
&= x\texttt{arctan}(2x+1)\:+ \frac{1}{2}\texttt{arctan}(2x+1)\:- \frac{1}{2}\int\frac{u}{u^2+1}du
& \color{gray}{\text{[Substitute x back to the recently integrated term]}} \\[5pt]
&= \texttt{arctan}(2x+1) \left[ x+\frac{1}{2} \right] \:- \frac{1}{2}\int\frac{u}{u^2+1}du
& \color{gray}{\text{[Factor the arctan]}} \\[5pt]
\end{align*}
\)

Use U-Sub with the given integrand. Let \(v=u^2\), so \(dv=2udu\). Substituting accordingly yields:

\(\normalsize
\begin{align*}\displaystyle
\texttt{arctan}(2x+1) \left[ x+\frac{1}{2} \right] \:- \frac{1}{2}\int\frac{u}{u^2+1}du
&= \texttt{arctan}(2x+1) \left[ x+\frac{1}{2} \right] \:- \frac{1}{2}\int\frac{\color{red}{\frac{du}{2}}}{\color{red}{v}+1}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \texttt{arctan}(2x+1) \left[ x+\frac{1}{2} \right] \:- \frac{1}{4}\int\frac{du}{v+1}
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= \texttt{arctan}(2x+1) \left[ x+\frac{1}{2} \right] \:- \frac{1}{4}\texttt{ln}|v+1|\:+ C
& \color{gray}{\text{[Integrate]}} \\[5pt]
&= \texttt{arctan}(2x+1) \left[ x+\frac{1}{2} \right] \:- \frac{1}{4}\texttt{ln}(u^2+1)\:+ C
& \color{gray}{\text{[Substitute back for v (ln argument always positive)]}} \\[5pt]
&= \color{green}{\texttt{arctan}(2x+1) \left[ x+\frac{1}{2} \right] \:- \frac{1}{4}\texttt{ln}((2x+1)^2+1)\:+C}
& \color{gray}{\text{[Substitute back for x]}} \\[5pt]
\end{align*}
\)

Problem 4: Solution

Let \(u=\texttt{cos}(\sqrt{x})\), so \(du=\frac{-\texttt{sin}(\sqrt{x})}{2\sqrt{x}}dx\); and \(dv=dx\), so \(v=x\). Substituting these values into \(uv – \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{cos}{\sqrt{x}}dx
&= \color{red}{\underbrace{\color{black}{\texttt{cos}(\sqrt{x})}}_{\large u}} \color{red}{\underbrace{\color{black}{x}}_{\large v}}\:- \int\ \color{red}{\underbrace{\color{black}{x}}_{\large v}} \color{red}{\underbrace{\color{black}{\left(\frac{-\texttt{sin}(\sqrt{x})}{2\sqrt{x}}\right)}}_{\large du}}dx
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{cos}(\sqrt{x})\:+ \frac{1}{2}\int\frac{x\texttt{sin}(\sqrt{x})}{\sqrt{x}}dx
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= x\texttt{cos}(\sqrt{x})\:+ \frac{1}{2}\int\sqrt{x}\texttt{sin}(\sqrt{x})dx
& \color{gray}{\text{[Simplify terms]}} \\[5pt]
\end{align*}
\)

Use U-Sub with the given integrand. Let \(u=\sqrt{x}\), so \(du=\frac{1}{2\sqrt{x}}dx\); \(dx=2\sqrt{x}du=2udu\). Substituting accordingly yields:

\(\normalsize
\begin{align*}\displaystyle
x\texttt{cos}(\sqrt{x})\:+ \frac{1}{2}\int\sqrt{x}\texttt{sin}(\sqrt{x})dx
&= x\texttt{cos}(\sqrt{x})\:+ \frac{1}{2}\int \color{red}{\underbrace{\color{black}{u}}_{\large u^2}} \texttt{sin}(\color{red}{\underbrace{\color{black}{u}}_{\large u}})\color{red}{(2udu)}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{cos}(\sqrt{x})\:+ \int u^2\texttt{sin}(u)du
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
\end{align*}
\)

Apply IBP again. Let \(m=u^2\), so \(dm=2udu\); and \(dw=\texttt{sin}(u)du\), so \(w=-\texttt{cos}(u)\). Substituting these values into \(mw\:- \int wdm\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
x\texttt{cos}(\sqrt{x})\:+ \int u^2\texttt{sin}(u)du
&= x\texttt{cos}(\sqrt{x})\:+ \left[ \left( \color{red}{\underbrace{\color{black}{u^2}}_{\large m}}\right) \left( \color{red}{\underbrace{\color{black}{-\texttt{cos}(u)}}_{\large w}} \right)\: -\int \left( \color{red}{\underbrace{\color{black}{-\texttt{cos}(u)}}_{\large w}}\right) \left( \color{red}{\underbrace{\color{black}{2udu}}_{\large dm}} \right) \right]
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{cos}(\sqrt{x})\:- u^2\texttt{cos}(u)\:+ 2\int u\texttt{cos}(u)du
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
\end{align*}
\)

Apply IBP again. Let \(m=u\), so \(dm=du\); and \(dw=\texttt{cos}(u)du\), so \(w=\texttt{sin}(u)\). Substituting these values into \(mw\:- \int wdm\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
x\texttt{cos}(\sqrt{x})\: &- u^2\texttt{cos}(u)\; + 2\int u\texttt{cos}(u)du\\[5pt]
&= \texttt{cos}(\sqrt{x})\:- u^2\texttt{cos}(u)\:+ 2\left( (\color{red}{\underbrace{\color{black}{u}}_{\large m}})(\color{red}{\underbrace{\color{black}{\texttt{sin}(u)}}_{\large w}})\:- \int \color{red}{\underbrace{\color{black}{\texttt{sin}(u)}}_{\large w}} \color{red}{\underbrace{\color{black}{du}}_{dm}} \right)
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{cos}(\sqrt{x})\:- u^2\texttt{cos}(u)\:+ 2u\texttt{sin}(u)\:- 2\int \texttt{sin}(u)du
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= x\texttt{cos}(\sqrt{x})\:- u^2\texttt{cos}(u)\:+ 2u\texttt{sin}(u)\:+ 2\texttt{cos}(u)\:+ C
& \color{gray}{\text{[Integrate]}} \\[5pt]
&= x\texttt{cos}(\sqrt{x})\:- x\texttt{cos}(\sqrt{x})\:+ 2\sqrt{x}\texttt{sin}(\sqrt{x})\:+ 2\texttt{cos}(\sqrt{x})\:+ C
& \color{gray}{\text{[Substitute back to x]}} \\[5pt]
&= \color{green}{2\sqrt{x}\texttt{sin}(\sqrt{x})\:+ 2\texttt{cos}(\sqrt{x})\:+ C}
& \color{gray}{\text{[Simplify]}} \\[5pt]
\end{align*}
\)

Problem 5: Solution

Let \(u=\texttt{e}^{\sqrt{x}}\), so \(du=\frac{\texttt{e}^{\sqrt{x}}}{2\sqrt{x}}dx\); and \(dv=dx\), so \(v=x\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{e}^{\sqrt{x}}dx
&= \color{red}{\underbrace{\color{black}{\left( \texttt{e}^{\sqrt{x}} \right)}}_{\large u}} \color{red}{\underbrace{\color{black}{x}}_{\large v}}\:- \int \color{red}{\underbrace{\color{black}{x}}_{\large v}} \color{red}{\underbrace{\color{black}{\left( \frac{\texttt{e}^{\sqrt{x}}}{2\sqrt{x}}dx \right)}}_{\large du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{e}^{\sqrt{x}}\:- \frac{1}{2}\int\frac{x\texttt{e}^{\sqrt{x}}}{\sqrt{x}}dx
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= x\texttt{e}^{\sqrt{x}}\:- \frac{1}{2}\int \sqrt{x}\texttt{e}^{\sqrt{x}}dx
& \color{gray}{\text{[Simplify]}} \\[5pt]
\end{align*}
\)

Use U-Sub with the given integrand. Let \(u=\sqrt{x}\), so \(du=\frac{1}{2\sqrt{x}}dx\); \(dx=2\sqrt{x}du=2udu\). Substituting accordingly yields:

\(\normalsize
\begin{align*}\displaystyle
x\texttt{e}^{\sqrt{x}}\:- \frac{1}{2}\int \sqrt{x}\texttt{e}^{\sqrt{x}}dx
&= x\texttt{e}^{\sqrt{x}}\:- \frac{1}{2}\int \color{red}{2u^2\texttt{e}^udu}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{e}^{\sqrt{x}}\:- \int u^2\texttt{e}^udu
& \color{gray}{\text{[Simplify]}} \\[5pt]
\end{align*}
\)

Apply IBP. Let \(m=u^2\), so \(dm=2udu\); and \(dw=\texttt{e}^udu\), so \(w=\texttt{e}^u\). Substituting these values into \(mw\:- \int wdm\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
x\texttt{e}^{\sqrt{x}}\:- \int u^2\texttt{e}^udu
&= x\texttt{e}^{\sqrt{x}}\:- \left( \color{red}{\underbrace{\color{black}{u^2}}_{\large m}} \color{red}{\underbrace{\color{black}{\texttt{e}^u}}_{\large w}}\:- \int \color{red}{\underbrace{\color{black}{\texttt{e}^u}}_{\large w}} \color{red}{\underbrace{\color{black}{(2udu)}}_{dm}} \right)
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{e}^{\sqrt{x}}\:- u^2\texttt{e}^u\:+ 2\int u\texttt{e}^udu
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
\end{align*}
\)

Apply IBP one last time. Let \(m=u\), so \(dm=du\); and \(dw=e^udu\), so \(w=e^u\). Substituting these values into \(mw\:- \int wdm\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
x\texttt{e}^{\sqrt{x}}\:- u^2\texttt{e}^u\:+ 2\int u\texttt{e}^udu
&= x\texttt{e}^{\sqrt{x}}\:- u^2\texttt{e}^u\:+ 2\left( \color{red}{\underbrace{\color{black}{u}}_{\large m}} \color{red}{\underbrace{\color{black}{\texttt{e}^u}}_{\large w}}\:- \int \color{red}{\underbrace{\color{black}{\texttt{e}^u}}_{\large w}} \color{red}{\underbrace{\color{black}{du}}_{dm}} \right)
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{e}^{\sqrt{x}}\:- u^2\texttt{e}^u\:+ 2u\texttt{e}^u\:- 2\int \texttt{e}^udu
& \color{gray}{\text{[Rearrange the expression]}} \\[5pt]
&= x\texttt{e}^{\sqrt{x}}\:- u^2\texttt{e}^u\:+ 2u\texttt{e}^u\:- 2\texttt{e}^u\:+ C
& \color{gray}{\text{[Intergate]}} \\[5pt]
&= x\texttt{e}^{\sqrt{x}}\:- x\texttt{e}^{\sqrt{x}}\:+ 2\sqrt{x}\texttt{e}^{\sqrt{x}}\:- 2\texttt{e}^{\sqrt{x}}\:+ C
& \color{gray}{\text{[Substitute back to x]}} \\[5pt]
&= 2\sqrt{x}\texttt{e}^{\sqrt{x}}\:- 2\texttt{e}^{\sqrt{x}}\:+ C
& \color{gray}{\text{[Simplify]}} \\[5pt]
&= \color{green}{2\texttt{e}^{\sqrt{x}}(\sqrt{x}-1)\:+ C}
& \color{gray}{\text{[Factor]}} \\[5pt]
\end{align*}
\)

Problem 6: Solution

Let \(u=\texttt{sin(ln(x))}\), so \(du=\frac{\texttt{cos(ln(x))}}{x}dx\); and \(dv=dx\), so \(v=x\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{sin(ln(x))}dx
&= \color{red}{\underbrace{\color{black}{x}}_{\large v}} \color{red}{\underbrace{\color{black}{\texttt{sin(ln(x))}}}_{\large u}}\:- \int \color{red}{\underbrace{\color{black}{x}}_{\large v}} \color{red}{\underbrace{\color{black}{\left( \frac{\texttt{cos(ln(x))}}{x}dx \right)}}_{\large du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{sin(ln(x))}\:- \int\texttt{cos(ln(x))} dx
& \color{gray}{\text{[Simplify]}} \\[5pt]
\end{align*}
\)

Use IBP again. Let \(u=\texttt{cos(ln(x))}\), so \(du=\frac{-\texttt{sin(ln(x))}}{x}dx\); and \(dv=dx\), so \(v=x\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
x\texttt{sin(ln(x))}\:- \int\texttt{cos(ln(x))} dx
&= x\texttt{sin(ln(x))}\:- \left( \color{red}{\underbrace{\color{black}{x}}_{\large v}} \color{red}{\underbrace{\color{black}{\texttt{cos(ln(x))}}}_{\large u}}\:- \int \color{red}{\underbrace{\color{black}{x}}_{\large v}} \color{red}{\underbrace{\color{black}{\left( \frac{-\texttt{sin(ln(x))}}{x}dx \right)}}_{\large du}} \right)
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= x\texttt{sin(ln(x))}\:- x\texttt{cos(ln(x))}\:- \int \texttt{sin(ln(x))}dx
& \color{gray}{\text{[Simplify]}}
\end{align*}
\)

Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:

\(\normalsize
\begin{align*}\displaystyle
& \color{red}{\int \texttt{sin(ln(x))}dx} = x\texttt{sin(ln(x))}\:- x\texttt{cos(ln(x))}\:- \color{red}{\int \texttt{sin(ln(x))}dx}
& \color{gray}{\text{[Form an equation with both integrals]}} \\[5pt]
& \color{red}{2\int \texttt{sin(ln(x))}dx} = x\texttt{sin(ln(x))}\:- x\texttt{cos(ln(x))}\:+ C
& \color{gray}{\text{[Arrange both integrals to the left hand side of the equation]}} \\[5pt]
& \color{green}{\int \texttt{sin(ln(x))}dx = \frac{x}{2} [\texttt{sin(ln(x))}\:- \texttt{cos(ln(x))}]\:+ C}
& \color{gray}{\text{[Divide both sides by 2 and factor x]}}
\end{align*}
\)

Problem 7: Solution

Let \(u=\texttt{cos}(5x)\), so \(du=-5\texttt{sin}(5x)dx\); and \(dv=\texttt{e}^{2x}dx\), so \(v=\frac{1}{2}\texttt{e}^{2x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{e}^{2x}\texttt{cos}(5x)dx
&= \color{red}{\underbrace{\color{black}{\frac{1}{2}\texttt{e}^{2x}}}_{\large v}} \color{red}{\underbrace{\color{black}{\texttt{cos}(5x)}}_{\large u}}\:- \int \color{red}{\underbrace{\color{black}{\left( \frac{1}{2}\texttt{e}^{2x} \right)}}_{\large v}} \color{red}{\underbrace{\color{black}{(-5\texttt{cos}(5x)dx)}}_{\large du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{2}\texttt{e}^{2x}\texttt{cos}(5x)\:+ \frac{5}{2}\int \texttt{e}^{2x}\texttt{sin}(5x)dx
& \color{gray}{\text{[Simplify]}} \\[5pt]
\end{align*}
\)

Use IBP again. Let \(u=\texttt{sin}(5x)\), so \(du=5\texttt{cos}(5x)dx\); and \(dv=\texttt{e}^{2x}dx\), so \(v=\frac{1}{2}\texttt{e}^{2x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\frac{1}{2}\texttt{e}^{2x}\texttt{cos}(5x)\:+ \frac{5}{2}\int \texttt{e}^{2x}\texttt{sin}(5x)dx
&=\frac{1}{2}\texttt{e}^{2x}\texttt{cos}(5x)\:+ \frac{5}{2} \left( \color{red}{\underbrace{\color{black}{\frac{1}{2}\texttt{e}^{2x}}}_{\large v}} \color{red}{\underbrace{\color{black}{\texttt{sin}(5x)}}_{\large u}}\:- \int \color{red}{\underbrace{\color{black}{\left( \frac{1}{2}\texttt{e}^{2x} \right)}}_{\large v}} \color{red}{\underbrace{\color{black}{(5\texttt{cos}(5x)dx)}}_{\large du}} \right)
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{2} \texttt{e}^{2x}\texttt{cos}(5x)\:+ \frac{5}{4}\texttt{e}^{2x}\texttt{sin}(5x)\:- \frac{25}{4}\int \texttt{e}^{2x}\texttt{cos}(5x)dx
& \color{gray}{\text{[Simplify]}}
\end{align*}
\)

\(\normalsize
\begin{align*}\displaystyle
& \color{red}{\int \texttt{e}^{2x}\texttt{cos}(5x)dx} = \frac{1}{2} \texttt{e}^{2x}\texttt{cos}(5x)\:+ \frac{5}{4}\texttt{e}^{2x}\texttt{sin}(5x)\:- \color{red}{\frac{25}{4}\int \texttt{e}^{2x}\texttt{cos}(5x)dx}
& \color{gray}{\text{[Form an equation with both integrals]}} \\[5pt]
& \color{red}{\frac{29}{4}\int \texttt{e}^{2x}\texttt{cos}(5x)dx} = \frac{1}{2} \texttt{e}^{2x}\texttt{cos}(5x)\:+ \frac{5}{4}\texttt{e}^{2x}\texttt{sin}(5x)\:+ C
& \color{gray}{\text{[Arrange both integrals to the left hand side of the equation]}} \\[5pt]
& \color{red}{\int \texttt{e}^{2x}\texttt{cos}(5x)dx} = \left( \frac{4}{29} \right) \left( \frac{1}{2}\texttt{e}^{2x} \right) (\texttt{cos}(5x)\:+ \frac{5}{2}\texttt{sin}(5x))\:+ C
& \color{gray}{\text{[Factor]}} \\[5pt]
& \color{green}{\int \texttt{e}^{2x}\texttt{cos}(5x)dx = \left( \frac{1}{29}\texttt{e}^{2x} \right)[2\texttt{cos}(5x)\:+ 5\texttt{sin}(5x)]\:+ C}
& \color{gray}{\text{[Simplify and distribute the numerator]}}
\end{align*}
\)

Problem 8: Solution

Let \(u=\texttt{sin}(\beta x)\), so \(du=\beta \texttt{cos}(\beta x)dx\); and \(dv=\texttt{e}^{\alpha x}dx\), so \(v=\frac{1}{\alpha}\texttt{e}^{\alpha x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx
&= \color{red}{\underbrace{\color{black}{\frac{1}{\alpha}\texttt{e}^{\alpha x}}}_{\large v}} \color{red}{\underbrace{\color{black}{\texttt{sin}(\beta x)}}_{\large u}}\:- \int \color{red}{\underbrace{\color{black}{\left( \frac{1}{\alpha}\texttt{e}^{\alpha x} \right)}}_{\large v}} \color{red}{\underbrace{\color{black}{(\beta \texttt{cos}(\beta x)dx)}}_{\large du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{\alpha}\texttt{e}^{\alpha x}\texttt{sin}(\beta x)\:- \frac{\beta}{\alpha}\int \texttt{e}^{\alpha x}\texttt{cos}(\beta x)dx
& \color{gray}{\text{[Simplify]}} \\[5pt]
\end{align*}
\)

Use IBP again. Let \(u=\texttt{cos}(\beta x)\), so \(du=-\beta \texttt{sin}(\beta x)dx\); and \(dv=\texttt{e}^{\alpha x}dx\), so \(v=\frac{1}{\alpha }\texttt{e}^{\alpha x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\frac{1}{\alpha}\texttt{e}^{\alpha x}\texttt{sin}(\beta x)\:- \frac{\beta}{\alpha}\int \texttt{e}^{\alpha x}\texttt{cos}(\beta x)dx
&=\frac{1}{\alpha}\texttt{e}^{\alpha x}\texttt{sin}(\beta x)\:- \frac{\beta}{\alpha} \left( \color{red}{\underbrace{\color{black}{\frac{1}{\alpha}\texttt{e}^{\alpha x}}}_{\large v}} \color{red}{\underbrace{\color{black}{\texttt{cos}(\beta x)}}_{\large u}}\:- \int \color{red}{\underbrace{\color{black}{\left( \frac{1}{\alpha}\texttt{e}^{\alpha x} \right)}}_{\large v}} \color{red}{\underbrace{\color{black}{(-\beta \texttt{sin}(\beta x)dx)}}_{\large du}} \right)
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{\alpha} \texttt{e}^{\alpha x}\texttt{sin}(\beta x)\:- \frac{\beta}{\alpha^2}\texttt{e}^{\alpha x}\texttt{cos}(\beta x)\:- \frac{\beta^2}{\alpha^2}\int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx
& \color{gray}{\text{[Simplify]}}
\end{align*}
\)

\(\normalsize
\begin{align*}\displaystyle
& \color{red}{\int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx} = \frac{1}{\alpha} \texttt{e}^{\alpha x}\texttt{sin}(\beta x)\:- \frac{\beta}{\alpha^2}\texttt{e}^{\alpha x}\texttt{cos}(\beta x)\:- \color{red}{\frac{\beta^2}{\alpha^2}\int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx}
& \color{gray}{\text{[Form an equation with both integrals]}} \\[5pt]
& \color{red}{\frac{\beta^2\:+ \alpha^2}{\alpha^2} \int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx} = \frac{1}{\alpha} \texttt{e}^{\alpha x}\texttt{sin}(\beta x)\:- \frac{\beta}{\alpha^2}\texttt{e}^{\alpha x}\texttt{cos}(\beta x)\:+ C
& \color{gray}{\text{[Arrange both integrals to the left hand side of the equation]}} \\[5pt]
& \color{red}{\int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx} = \left( \frac{\alpha^2}{\alpha^2+\beta^2} \right) \left( \frac{1}{\alpha}\texttt{e}^{\alpha x} \right) [\texttt{sin}(\beta x)\:- \frac{\beta}{\alpha}\texttt{cos}(\beta x)]\:+ C
& \color{gray}{\text{[Factor]}} \\[5pt]
& \int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx = \left( \frac{\alpha}{\alpha^2+\beta^2} \right) \left(\texttt{e}^{\alpha x} \right) [\texttt{sin}(\beta x)\:- \frac{\beta}{\alpha}\texttt{cos}(\beta x)]\:+ C
& \color{gray}{\text{[Simplify]}} \\[5pt]
& \color{green}{\int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx = \left( \frac{\texttt{e}^{\alpha x}}{\alpha^2\:+ \beta^2} \right)[\alpha\texttt{sin}(\beta x)\:- \beta\texttt{cos}(\beta x)]\:+ C}
& \color{gray}{\text{[Distribute the numerator]}}
\end{align*}
\)

Problem 9: Solution

Let \(u=\texttt{cos}(\beta x)\), so \(du=-\beta \texttt{sin}(\beta x)dx\); and \(dv=\texttt{e}^{\alpha x}dx\), so \(v=\frac{1}{\alpha}\texttt{e}^{\alpha x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\int \texttt{e}^{\alpha x}\texttt{cos}(\beta x)dx
&= \color{red}{\underbrace{\color{black}{\frac{1}{\alpha}\texttt{e}^{\alpha x}}}_{\large v}} \color{red}{\underbrace{\color{black}{\texttt{cos}(\beta x)}}_{\large u}}\:- \int \color{red}{\underbrace{\color{black}{\left( \frac{1}{\alpha}\texttt{e}^{\alpha x} \right)}}_{\large v}} \color{red}{\underbrace{\color{black}{(-\beta \texttt{sin}(\beta x)dx)}}_{\large du}}
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{\alpha}\texttt{e}^{\alpha x}\texttt{cos}(\beta x)\:+ \frac{\beta}{\alpha}\int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx
& \color{gray}{\text{[Simplify]}} \\[5pt]
\end{align*}
\)

Use IBP again. Let \(u=\texttt{sin}(\beta x)\), so \(du=\beta \texttt{cos}(\beta x)dx\); and \(dv=\texttt{e}^{\alpha x}dx\), so \(v=\frac{1}{\alpha }\texttt{e}^{\alpha x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:

\(\normalsize
\begin{align*}\displaystyle
\frac{1}{\alpha}\texttt{e}^{\alpha x}\texttt{cos}(\beta x)\:+ \frac{\beta}{\alpha}\int \texttt{e}^{\alpha x}\texttt{sin}(\beta x)dx
&=\frac{1}{\alpha}\texttt{e}^{\alpha x}\texttt{cos}(\beta x)\:+ \frac{\beta}{\alpha} \left( \color{red}{\underbrace{\color{black}{\frac{1}{\alpha}\texttt{e}^{\alpha x}}}_{\large v}} \color{red}{\underbrace{\color{black}{\texttt{sin}(\beta x)}}_{\large u}}\:- \int \color{red}{\underbrace{\color{black}{\left( \frac{1}{\alpha}\texttt{e}^{\alpha x} \right)}}_{\large v}} \color{red}{\underbrace{\color{black}{(\beta \texttt{cos}(\beta x)dx)}}_{\large du}} \right)
& \color{gray}{\text{[Apply the substitution]}} \\[5pt]
&= \frac{1}{\alpha} \texttt{e}^{\alpha x}\texttt{cos}(\beta x)\:+ \frac{\beta}{\alpha^2}\texttt{e}^{\alpha x}\texttt{sin}(\beta x)\:- \frac{\beta^2}{\alpha^2}\int \texttt{e}^{\alpha x}\texttt{cos}(\beta x)dx
& \color{gray}{\text{[Simplify]}}
\end{align*}
\)

\(\normalsize
\begin{align*}\displaystyle
& \color{red}{\int \texttt{e}^{\alpha x}\texttt{cos}(\beta x)dx} = \frac{1}{\alpha} \texttt{e}^{\alpha x}\texttt{cos}(\beta x)\:+ \frac{\beta}{\alpha^2}\texttt{e}^{\alpha x}\texttt{sin}(\beta x)\:- \color{red}{\frac{\beta^2}{\alpha^2}\int \texttt{e}^{\alpha x}\texttt{cos}(\beta x)dx}
& \color{gray}{\text{[Form an equation with both integrals]}} \\[5pt]
& \color{red}{\frac{\beta^2\:+ \alpha^2}{\alpha^2} \int \texttt{e}^{\alpha x}\texttt{cos}(\beta x)dx} = \frac{1}{\alpha} \texttt{e}^{\alpha x}\texttt{cos}(\beta x)\:+ \frac{\beta}{\alpha^2}\texttt{e}^{\alpha x}\texttt{sin}(\beta x)\:+ C
& \color{gray}{\text{[Arrange both integrals to the left hand side of the equation]}} \\[5pt]
& \color{red}{\int \texttt{e}^{\alpha x}\texttt{cos}(\beta x)dx} = \left( \frac{\alpha^2}{\alpha^2+\beta^2} \right) \left( \frac{1}{\alpha}\texttt{e}^{\alpha x} \right) [\texttt{cos}(\beta x)\:+ \frac{\beta}{\alpha}\texttt{sin}(\beta x)]\:+ C
& \color{gray}{\text{[Factor]}} \\[5pt]
& \int \texttt{e}^{\alpha x}\texttt{cos}(\beta x)dx = \left( \frac{\alpha}{\alpha^2+\beta^2} \right) \left(\texttt{e}^{\alpha x} \right) [\texttt{cos}(\beta x)\:+ \frac{\beta}{\alpha}\texttt{sin}(\beta x)]\:+ C
& \color{gray}{\text{[Simplify]}} \\[5pt]
& \color{green}{\int \texttt{e}^{\alpha x}\texttt{cos}(\beta x)dx = \left( \frac{\texttt{e}^{\alpha x}}{\alpha^2\:+ \beta^2} \right)[\alpha\texttt{cos}(\beta x)\:+ \beta\texttt{sin}(\beta x)]\:+ C}
& \color{gray}{\text{[Distribute the numerator]}}
\end{align*}
\)

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